I was wrong because in this examples p was a pointer to int. Sorry, I was
thinking on something like this:

int *p;
*p = 0;

Cheers,
Emiliano.

On Mon, Mar 28, 2016 at 2:16 PM, Emiliano Marini <emilianomarini82@???
> wrote:

> You're right.
>
> On Mon, Mar 28, 2016 at 2:04 PM, Rainer Weikusat <
> rainerweikusat@???> wrote:
>
>> Emiliano Marini <emilianomarini82@???> writes:
>> > char *p;
>> > p="01234"; /* skeezy, but makes the point */
>> >
>> > Warning! Here "p" is pointing to nowhere, you don't know which memory
>> > locations are writing to.
>>
>> The 'memory location' (if any) reserved for the pointer p itself by the
>> compiler, IOW, this is totally correct.
>>
>> >
>> > char *p;
>> > *p=malloc...*
>> > p="01234"; /* skeezy, but makes the point */
>>
>> And this is a memory leak as the pointer returned by malloc is
>> overwritten.
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>